A body moves on a frictionless plane starting from rest. If S_n is distance moved between t=n-1 and t=n and S_ n-1 is distance moved between t=n-2 and t=n-1 , then the ratio S_ n-1 ~S_ n is (1-(2)/(x)) for n=10 . The

Correct answer: 0 aligned & S_n=(1)/(2) a(2 n-1)=(19 a)/(2) & S_ n-1 =(1)/(2) a(2 n-3)=(17 a)/(2) & S_ n-1 S_n =(17)/(19)=1-(2)/(x) x=19 aligned

JEE Main 20245 AprilMorning Shift

JEE Main 2024 · 5 April · Morning Shift · Physics · Motion in One Dimension

A body moves on a frictionless plane starting from rest. If S_n is...

A body moves on a frictionless plane starting from rest. If

S_n

is distance moved between t=n-1 and

\mathrm{t}=\mathrm{n} and \mathrm{S}_{\mathrm{n}-1}

is distance moved between

\mathrm{t}=\mathrm{n}-2

and

\mathrm{t}=\mathrm{n}-1

, then the ratio

\frac{\mathrm{S}_{\mathrm{n}-1}}{\mathrm{~S}_{\mathrm{n}}}

\left(1-\frac{2}{x}\right)

for

\mathrm{n}=10

. The value of

x

is ______.

\begin{aligned} & S_n=\frac{1}{2} a(2 n-1)=\frac{19 a}{2} \\ & S_{n-1}=\frac{1}{2} a(2 n-3)=\frac{17 a}{2} \\ & \frac{S_{n-1}}{S_n}=\frac{17}{19}=1-\frac{2}{x} \Rightarrow x=19\end{aligned}